# How to find the inverse of a function

Kupkake303
moved into h/w help, so template is missing
T(t) = Ts+(98.6 – Ts)e-kt

rewrite in the form t=g-1(T)

In trying to understand how to find the inverse of this but am having a hard time, please advise.

Thanks,
Kupkake303

Last edited by a moderator:

## Answers and Replies

Staff Emeritus
This is just a matter of trying to get $t$ by itself on one side of an equality. You initially have: $T = T_s + (98.6 - T_s) e^{-kt}$. To simplify the expression, let $U = e^{-kt}$. Then the equation is: $T = T_s + (98.6 - T_s) U$. So solve for $U$. Then solve for $t$ in terms of this value of $U$.

Kupkake303
This is just a matter of trying to get $t$ by itself on one side of an equality. You initially have: $T = T_s + (98.6 - T_s) e^{-kt}$. To simplify the expression, let $U = e^{-kt}$. Then the equation is: $T = T_s + (98.6 - T_s) U$. So solve for $U$. Then solve for $t$ in terms of this value of $U$.

So I would put it as 0=Ts+(98.6-Ts)U ??

Doing that would give me -Ts/(98.6-Ts)=U

Then putting the e-kt back in you get -Ts/(98.6-Ts)=e-kt

So then I would then need to get that by it's self so the t is by itself but I can't remember how to do this. Please advise. Thank you

Staff Emeritus
So I would put it as 0=Ts+(98.6-Ts)U ??

No, $T = T_s + (98.6 - T_s)U$

So $U = (T-T_s)/(98.6 - T_s)$

Kupkake303
No, $T = T_s + (98.6 - T_s)U$

So $U = (T-T_s)/(98.6 - T_s)$

Right, because the T wouldn't just disappear.

So next I would change the U back to a e-kt for e-kt=(T-Ts)/(98.6-Ts)

from there would I try to figure out the e-kt ?

Staff Emeritus
Right, because the T wouldn't just disappear.

So next I would change the U back to a e-kt for e-kt=(T-Ts)/(98.6-Ts)

from there would I try to figure out the e-kt ?

You want to take the natural log of both sides of the equation.

Kupkake303
You want to take the natural log of both sides of the equation.

okay so I would
e-kt=(T-Ts)/(98.6-Ts)

lne-kt=ln(T-Ts)/(98.6-Ts)

The ln would cancel out the e

-kt=ln(T-Ts)/(98.6-Ts)

then divide by -k? to get

t=(1/-k)ln(T-Ts)/(98.6-Ts)....?

so this would be my answer for finding the inverse ..?

Staff Emeritus
okay so I would
e-kt=(T-Ts)/(98.6-Ts)

lne-kt=ln(T-Ts)/(98.6-Ts)

The ln would cancel out the e

-kt=ln(T-Ts)/(98.6-Ts)

then divide by -k? to get

t=(1/-k)ln(T-Ts)/(98.6-Ts)....?

so this would be my answer for finding the inverse ..?

Yes, except that $-ln(A/B) = ln(B/A)$, so you can get rid of the $-$ sign to get:

$t = \frac{1}{k} ln((98.6 - T_s)/(T-T_s))$

Kupkake303
Yes, except that $-ln(A/B) = ln(B/A)$, so you can get rid of the $-$ sign to get:

$t = \frac{1}{k} ln((98.6 - T_s)/(T-T_s))$

Awesome! Thank you soooo much. I spent about 4-5 hours trying to figure this problem out myself and I wouldn't have gotten through it by myself.

Homework Helper
Gold Member
Logarithms are the inverse of the finish of exponentiation, by definition. Hopefully you remember your first lesson about them. Well hopefully they told you that then.